I think your post contains a lot of the right ideas, but as in my comment, the situation is much more clear if you allow field extensions.
Here is a positive result:
Lemma. Let $S$ be a scheme and let $X \to S$ and $Y \to S$ be finitely presented $S$-schemes. Let $k$ be a field and $s \colon \operatorname{Spec} k \to S$ a morphism, and let $\phi \colon s^*X \stackrel\sim\to s^*Y$ be an isomorphism. Then there exists a finite type morphism $f \colon T \to S$ with a point $t \colon \operatorname{Spec} k \to T$ such that $f \circ t = s$ and an isomorphism $\psi \colon f^* X \stackrel\sim\to f^*Y$ such that $\phi = t^*\psi$.
Proof. We may replace $S$ by an open neighbourhood of the image of $s$, so in particular we may assume $S$ is affine; say $S = \operatorname{Spec} R$. We can write $k$ as a filtered colimit of finite type $R$-algebras; in other words, we write $\operatorname{Spec} k \to S$ as a cofiltered limit of finite type affine morphisms $T_i \to S$. Then [Tag 01ZM] gives the result. $\square$
Corollary. Let $S$ be an integral scheme, let $k$ be a field, let $s \colon \operatorname{Spec} k \to S$ a morphism whose image is the generic point, and let $X \to S$ and $Y \to S$ be finitely presented $S$-schemes. If $s^*X \cong s^*Y$, then there exists a dense open subset $U \subseteq S$ such that $X_{\bar u} \cong Y_{\bar u}$ for all $u \in U$.
Recall that $X_{\bar u}$ means the base change of $X \to S$ along $\operatorname{Spec}\bigl(\ \!\overline{\kappa(u)}\ \!\bigr) \to S$.
Proof. Choose $f \colon T \to S$ as in the lemma. The image of $f$ contains a dense open subset $U$ since it contains the generic point and $f$ is of finite type. Moreover, for any $u \in U$ and any closed point $t \in T_u$, the residue field extension $\kappa(u) \to \kappa(t)$ is finite by the weak Nullstellensatz. Thus the lemma gives an isomorphism $X_{\kappa(t)} \cong Y_{\kappa(t)}$, so in particular when passing to the algebraic closure. $\square$
There might be a model theoretic proof in the case $S = \operatorname{Spec} \mathbf Z$, using that the theory of algebraically closed fields is complete. At any rate, if this proof exists, it is not really different from the argument above.
Remark. In general, it is certainly too much to expect that the isomorphism descends all the way to $\mathbf Z$. For instance you can take $X = Y = \mathbf A^1$ and $\phi \colon \mathbf A^1_{\mathbf C} \stackrel\sim\to \mathbf A^1_{\mathbf C}$ a translation by a transcendental number.
For the question of whether $\operatorname{Isom}_{\kappa(u)}(X_u,Y_u) \neq \varnothing$ without field extensions, we will stop keeping track of the precise isomorphism (although in the end it will pop up again).
Replace $S$ and $T$ in the conclusion of the lemma by the closures of the images of $s$ and $t$ to assume they are irreducible and $s$ and $t$ are their generic points. Pass to affine open neighbourhoods if you haven't already. Take a closed point in the generic fibre $T_s$, and replace $T$ by its closure to assume that $T \to S$ is generically finite (it is at this point that we lose the relationship between $\psi \colon f^*X \stackrel\sim\to f^*Y$ and the original isomorphism $\phi$, as $T$ no longer contains $t$).
Replace $S$ and $T$ by their underlying reduced schemes. If $\operatorname{char} \kappa(s) = 0$, then $T \to S$ is generically étale, so over some nonempty open $U \subseteq S$ it is finite étale. Replacing $S$ by $U$ we reduce to the case where $f \colon T \to S$ is finite étale, and finally we replace this finite étale cover by its Galois closure.
We still have an isomorphism $\psi \colon f^*X \stackrel\sim\to f^*Y$. This induces a 1-cocycle for $\operatorname{Gal}(T/S)$ in the sheaf $\mathscr Aut(X)$, so we find a Galois obstruction to the existence of a global isomorphism. Then the question of whether $X_u \cong Y_u$ for $u \in S$ becomes a question of Frobenius elements acting trivially on the cover $T \to S$.
Example. Let $X \subseteq \mathbf A^2$ be the curve $y^2 = f(x)$ for a monic cubic polynomial $f \in \mathbf Z[x]$, and $Y$ its quadratic twist $dy^2 = f(x)$. Their projective closures $\bar X$ and $\bar Y$ are elliptic curves with neutral element the unique point $O = [0:1:0]$ at infinity. They become isomorphic over $\mathbf Q[\sqrt{d}]$, but are not isomorphic over $\mathbf Q$. The latter follows from a coordinate calculation that is often omitted:
- Any isomorphism between elliptic curves in Weierstrass form is given by a projective change of coordinates;
- Table 1.2 in Chapter III of Silverman's The arithmetic of elliptic curves shows what a change of coordinates does to the equations;
- We may assume $X$ is given by $y^2 = x^3+ax+b$; then the quadratic twist can also be given by $y^2 = x^3 + d^2ax + d^3b$;
- Then the only change of coordinates that is possible (in the notation of Silverman) has $r=s=t=0$ and $u$ needs to satisfy $u^2 = d$, which is impossible if $d$ is not a square.
If $p \nmid 6d\Delta(f)$ is a prime of good reduction for $X$ and for $\mathbf Z[\sqrt d]$, then $X_{\mathbf F_p} \cong Y_{\mathbf F_p}$ if and only if $p$ splits in $\mathbf Z[\sqrt{d}]$. Indeed, clearly the reductions at split primes become isomorphic, and at nonsplit primes we again get a quadratic twist, which for good primes $p > 3$ works the same as above. For instance, if $d = -1$ this happens if and only if $p \equiv 1 \pmod 4$.
Remark. What is a little weird about my method is that it's trying to descend a particular isomorphism. But there may exist some other isomorphism:
Example. Let $X = \mathbf P^1$ and $Y = V(x^2+y^2+z^2) \subseteq \mathbf P^2$. They become isomorphic over $\mathbf Q(i)$, but not over $\mathbf Q$ as $Y$ does not have a rational point. But they are isomorphic over $\mathbf F_p$ for every prime $p > 2$! Indeed, by Chevalley–Warning, the equation $x^2+y^2+z^2=0$ has a nontrivial solution in $\mathbf F_p$, and any smooth genus $0$ curve with a rational point is isomorphic to $\mathbf P^1$. However, we cannot choose the isomorphism "uniformly" in $p$ as it does not come from $\mathbf Z$.
Remark. This type of question is widely studied in arithmetic algebraic geometry, and there seems little hope of a clean criterion when $X_{\mathbf F_p} \cong Y_{\mathbf F_p}$ for all primes $p$, almost all primes $p$, a density $1$ set of primes $p$, a positive density set of primes $p$, etcetera.
Another widely studied question is the same but with $\mathbf F_p$ replaced by $\mathbf Q_p$; for instance the genus $1$ curve $3x^3+4y^3+5z^3=0$ in $\mathbf P^2$ is not isomorphic to its Jacobian over $\mathbf Q$ as it does not have a rational point, but they do become isomorphic over $\mathbf Q_p$ for all $p$ (including $\mathbf Q_\infty = \mathbf R$). This is a first failure of a 'local-to-global principle' (this example is due to Selmer, and launched the study of what is now called the Selmer group).
